Question

जर tanθ = 1 तर `(sinθ + cosθ)/(secθ + "cosec"θ)` ची किंमत काढा.

Answer 1

tan θ = 1 .....…[दिलेले]

आपल्याला माहीत आहे, की tan45° = 1

∴ tan θ = tan45°

∴ θ = 45° 

आता, sin θ = sin45° = `1/sqrt2`

cos θ = cos45° = `1/sqrt2`

sec θ = sec45° = `sqrt2`

`"cosec" θ = "cosec"45°` = `sqrt2`

`(sin θ + cos θ)/(sec θ + "cosec" θ) = (1/sqrt2 + 1/sqrt2)/(sqrt2 + sqrt2)`

= `((2/sqrt2))/(2sqrt2) = 2/sqrt2 ÷ 2sqrt2`

= `2/sqrt2 xx 1/(2sqrt2)`

∴ `(sin θ + cos θ)/(sec θ + "cosec" θ) = 1/2`